Integrand size = 33, antiderivative size = 610 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a^3 \left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{11/2} f}+\frac {a^3 \left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{11/2} f}-\frac {2 a^3 g (g \cos (e+f x))^{3/2}}{3 b^4 f}+\frac {2 a (g \cos (e+f x))^{7/2}}{7 b^2 f g}-\frac {2 a^4 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^5 f \sqrt {\cos (e+f x)}}+\frac {6 a^2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {4 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 b f \sqrt {\cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^6 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^4 \left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^6 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a^2 g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^3 f}+\frac {4 g (g \cos (e+f x))^{3/2} \sin (e+f x)}{45 b f}-\frac {2 (g \cos (e+f x))^{7/2} \sin (e+f x)}{9 b f g} \]
-a^3*(-a^2+b^2)^(3/4)*g^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^ 2)^(1/4)/g^(1/2))/b^(11/2)/f+a^3*(-a^2+b^2)^(3/4)*g^(5/2)*arctanh(b^(1/2)* (g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(11/2)/f-2/3*a^3*g*(g*cos (f*x+e))^(3/2)/b^4/f+2/7*a*(g*cos(f*x+e))^(7/2)/b^2/f/g+2/5*a^2*g*(g*cos(f *x+e))^(3/2)*sin(f*x+e)/b^3/f+4/45*g*(g*cos(f*x+e))^(3/2)*sin(f*x+e)/b/f-2 /9*(g*cos(f*x+e))^(7/2)*sin(f*x+e)/b/f/g+a^4*(a^2-b^2)*g^3*(cos(1/2*f*x+1/ 2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^ 2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^6/f/(b-(-a^2+b^2)^(1/2))/(g*cos( f*x+e))^(1/2)+a^4*(a^2-b^2)*g^3*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1 /2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos( f*x+e)^(1/2)/b^6/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-2*a^4*g^2*(co s(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e), 2^(1/2))*(g*cos(f*x+e))^(1/2)/b^5/f/cos(f*x+e)^(1/2)+6/5*a^2*g^2*(cos(1/2* f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2 ))*(g*cos(f*x+e))^(1/2)/b^3/f/cos(f*x+e)^(1/2)+4/15*g^2*(cos(1/2*f*x+1/2*e )^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(g*cos (f*x+e))^(1/2)/b/f/cos(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 19.23 (sec) , antiderivative size = 790, normalized size of antiderivative = 1.30 \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{5/2} \left (\frac {\sin (e+f x) \left (-\frac {\left (15 a^4-9 a^2 b^2-2 b^4\right ) \csc (e+f x) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right )}{a^2-b^2}+\frac {(2+2 i) b^2 \left (-3 a^3+a b^2\right ) \left ((4-4 i) a \sqrt {b} \sqrt [4]{-a^2+b^2} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \left (a^2-b^2\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )\right )}{\left (-a^2+b^2\right )^{5/4} \sqrt {\sin ^2(e+f x)}}\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 b^{11/2} (a+b \sin (e+f x))}+\frac {\cos ^{\frac {3}{2}}(e+f x) \left (90 a b^2 \cos (2 (e+f x))+21 b \left (12 a^2+b^2\right ) \sin (e+f x)-5 \left (84 a^3-18 a b^2+7 b^3 \sin (3 (e+f x))\right )\right )}{42 b^4}\right )}{15 f \cos ^{\frac {5}{2}}(e+f x)} \]
((g*Cos[e + f*x])^(5/2)*((Sin[e + f*x]*(-(((15*a^4 - 9*a^2*b^2 - 2*b^4)*Cs c[e + f*x]*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos [e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3 /4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - L og[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4 )*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]])))/(a^2 - b^2)) + ((2 + 2*I)*b^2*(- 3*a^3 + a*b^2)*((4 - 4*I)*a*Sqrt[b]*(-a^2 + b^2)^(1/4)*AppellF1[3/4, 1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*x]^(3 /2) + 3*(a^2 - b^2)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a ^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4) *Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*S qrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]])))/((-a^2 + b^2)^(5/4)*Sqrt[Sin[e + f*x]^2]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(12*b^( 11/2)*(a + b*Sin[e + f*x])) + (Cos[e + f*x]^(3/2)*(90*a*b^2*Cos[2*(e + f*x )] + 21*b*(12*a^2 + b^2)*Sin[e + f*x] - 5*(84*a^3 - 18*a*b^2 + 7*b^3*Sin[3 *(e + f*x)])))/(42*b^4)))/(15*f*Cos[e + f*x]^(5/2))
Time = 1.55 (sec) , antiderivative size = 610, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sin ^3(e+f x) (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3 (g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (-\frac {a^3 (g \cos (e+f x))^{5/2}}{b^3 (a+b \sin (e+f x))}+\frac {a^2 (g \cos (e+f x))^{5/2}}{b^3}-\frac {a \sin (e+f x) (g \cos (e+f x))^{5/2}}{b^2}+\frac {\sin ^2(e+f x) (g \cos (e+f x))^{5/2}}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^4 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^5 f \sqrt {\cos (e+f x)}}-\frac {2 a^3 g (g \cos (e+f x))^{3/2}}{3 b^4 f}+\frac {6 a^2 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {2 a^2 g \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^3 f}+\frac {a^4 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^6 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^4 g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^6 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 g^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{11/2} f}+\frac {a^3 g^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{11/2} f}+\frac {2 a (g \cos (e+f x))^{7/2}}{7 b^2 f g}+\frac {4 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{15 b f \sqrt {\cos (e+f x)}}-\frac {2 \sin (e+f x) (g \cos (e+f x))^{7/2}}{9 b f g}+\frac {4 g \sin (e+f x) (g \cos (e+f x))^{3/2}}{45 b f}\) |
-((a^3*(-a^2 + b^2)^(3/4)*g^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/(( -a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(11/2)*f)) + (a^3*(-a^2 + b^2)^(3/4)*g^(5/ 2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/( b^(11/2)*f) - (2*a^3*g*(g*Cos[e + f*x])^(3/2))/(3*b^4*f) + (2*a*(g*Cos[e + f*x])^(7/2))/(7*b^2*f*g) - (2*a^4*g^2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^5*f*Sqrt[Cos[e + f*x]]) + (6*a^2*g^2*Sqrt[g*Cos[e + f*x]]* EllipticE[(e + f*x)/2, 2])/(5*b^3*f*Sqrt[Cos[e + f*x]]) + (4*g^2*Sqrt[g*Co s[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(15*b*f*Sqrt[Cos[e + f*x]]) + (a^4* (a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]) , (e + f*x)/2, 2])/(b^6*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + ( a^4*(a^2 - b^2)*g^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b ^2]), (e + f*x)/2, 2])/(b^6*(b + Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) + (2*a^2*g*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(5*b^3*f) + (4*g*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(45*b*f) - (2*(g*Cos[e + f*x])^(7/2)*Sin[e + f*x])/(9*b*f*g)
3.14.83.3.1 Defintions of rubi rules used
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 24.72 (sec) , antiderivative size = 2001, normalized size of antiderivative = 3.28
(-16*g^3*a*(1/168/b^4/g*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(24*sin(1/2*f* x+1/2*e)^6*b^2-36*sin(1/2*f*x+1/2*e)^4*b^2-14*sin(1/2*f*x+1/2*e)^2*a^2+18* sin(1/2*f*x+1/2*e)^2*b^2+7*a^2-3*b^2)-1/64/b^6*a^2*(a^2-b^2)/(g^2*(a^2-b^2 )/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^( 1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)) /(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/ 2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*g *cos(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b ^2)^(1/4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-(g^2*(a^2- b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))))+1/420*(g*(2*cos(1/2*f*x+1/2* e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*(1680*EllipticF(cos(1/2*f*x+1/2*e) ,2^(1/2))*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*si n(1/2*f*x+1/2*e)^2*a^4*b^2-2240*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin (1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2* e)^2*a^2*b^4+1680*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*x+1/2*e )^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*sin(1/2*f*x+1/2*e)^2*a^2*b^4-4 864*cos(1/2*f*x+1/2*e)^5*b^6+2240*cos(1/2*f*x+1/2*e)^5*a^2*b^4-3360*cos(1/ 2*f*x+1/2*e)^3*a^2*b^4+1008*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2 *f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)*b^6-1120*cos(1/2*f*x +1/2*e)*sin(1/2*f*x+1/2*e)^2*a^2*b^4-400*EllipticF(cos(1/2*f*x+1/2*e),2...
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \sin ^3(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]